In a certain sequence the first term is $a_1 = 2007$ and the second term is $a_2 = 2008.$  Furthermore, the values of the remaining terms are chosen so that
\[a_n + a_{n + 1} + a_{n + 2} = n\]for all $n \ge 1.$  Determine $a_{1000}.$
Solution: We know that $a_n + a_{n + 1} + a_{n + 2} = n$ and $a_{n - 1} + a_n + a_{n + 1} = n - 1.$  Subtracting these equations, we get
\[a_{n + 2} - a_{n - 1} = 1,\]so $a_{n + 2} = a_{n - 1} + 1.$

Hence, the terms
\[a_1 = 2007, \ a_4, \ a_7, \ a_{10}, \ \dots, \ a_{1000}\]form an arithmetic sequence with common difference 1.  The common difference of 1 is added $\frac{1000 - 1}{3} = 333$ times, so $a_{1000} = 2007 + 333 = \boxed{2340}.$